On the second part of Hilbert s 16th problem

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1 Nonlinear Analysis ( ) On the second part of Hilbert s 6th problem Elin Oxenhielm Department of Mathematics, Stockholm University, Stockholm 69, Sweden Received 3 July 23; accepted 3 October 23 Abstract Let k be an integer such that k is larger than or equal to zero, and let H be the Hilbert number. In this paper, we use the method of describing functions to prove that in the Lienard equation, the upper bound for H(2k ) is k. By applying this method to any planar polynomial vector eld, it is possible to completely solve the second part of Hilbert s 6th problem.? 23 Elsevier Ltd. All rights reserved. Keywords: Second part of Hilbert s 6th problem; Hilbert number; Lienard equation; Describing function; Limit cycle; Polynomial vector eld. Introduction In 9, Hilbert presented a list consisting of 23 mathematical problems (see []). The second part of the 6th problem appears to be one of the most persistent in that list, second only to the 8th problem, the Riemann conjecture. The second part of the 6th problem is traditionally split into three parts (see [2]). Problem. A limit cycle is an isolated closed orbit. Is it true that a planar polynomial vector eld has but a nite number of limit cycles? Problem 2. Is it true that the number of limit cycles of a planar polynomial vector eld is bounded by a constant depending on the degree of the polynomials only? Denote the degree of the planar polynomial vector eld by n. The bound on the number of limit cycles in Problem 2 is denoted by H(n), and is known as the Hilbert number. Linear vector elds have no limit cycles, hence H()=. Tel.: ; fax: address: (E. Oxenhielm). URL: X/$ - see front matter? 23 Elsevier Ltd. All rights reserved. doi:.6/j.na.23..2

2 2 E. Oxenhielm / Nonlinear Analysis ( ) Problem 3. Give an upper bound for H(n). Let k be an integer such that k. In 977, Lins et al. [3] found examples with k dierent limit cycles in the Lienard equation ẋ = y F(x); where ẏ = x; F(x)=q 2k x 2k q 2k x 2k q 2 x 2 q x: () The degree of this polynomial vector eld is denoted by 2k. The coecients q i (for integers i such that 6 i 6 2k) are real constants. Lins et al. [3] conjectured the number k as the upper bound for the number of limit cycles of the Lienard equation (). Their conjecture thus states that in the Lienard equation (), the upper bound for H(2k ) is k. In his list of mathematical problems for the next century, published in 998, Smale [4] mentioned the Lienard equation () as a simplied version of the second part of Hilbert s 6th problem (see [4]). In the present paper, we will prove the conjecture stated by Lins et al. [3] in 977, thereby solving the simplied version of the second part of Hilbert s 6th problem stated by Smale [4] in Preliminaries In this section, we will introduce the method of describing functions, which may be used to calculate limit cycles in nonlinear dynamic systems (see [4]). Consider a dynamic system ẋ = Mx h(x); where x is the m-dimensional vector of state variables, M is an m m constant matrix and h(x) is an m-dimensional vector of nonlinear functions. Assume that the state variables are dominated by a harmonic term of a specic order x = a a sin(!t); where a is the m-dimensional vector of center values, a is the m-dimensional vector of amplitudes and! is the frequency. a ; a and! are assumed to be real. a and! are nonzero. Then, approximate the vector of nonlinear functions by discarding higher harmonic terms (terms of the form cos(r!t) and sin(r!t) for integers r such that r 2) h(x) = Na sin(!t); where is an m-dimensional constant vector and N is an m m constant matrix. The components of N are called describing functions.

3 E. Oxenhielm / Nonlinear Analysis ( ) 3 The system becomes ẋ = Ma (M N)a sin(!t) and solutions for a ; a and! satisfy Ma =; (2) det(j!i M N)=; (3) where I is the m m identity matrix and j satises the equation j 2 =. 3. Result In this section, we will prove the conjecture stated by Lins et al. [3] in 977, by applying the method of describing functions to the Lienard equation (). Theorem. Let k be an integer such that k, and let H be the Hilbert number. For the Lienard equation (), we have that the upper bound for H(2k ) is k. Proof. Noticing that the state variable x of the Lienard equation () behaves approximately like a sine function in simulations (see Fig. ), we assume in order to make a good approximation of x that both state variables are dominated by a harmonic term of a specic order [ ] [ ] [ ] x a a = sin(!t) (4) y b b which gives that the Lienard equation () becomes [ ] [ ] [ ][ ] [ ][ ] ẋ f(t) q a q a = = sin(!t) ẏ g(t) b b [ ] q2k [a a sin(!t)] 2k q 2k [a a sin(!t)] 2k q 2 [a a sin(!t)] 2 : Since the nonlinear function only aects f(t), Eq. (2) gives that the constant part of g(t) may be set equal to zero at this stage, so that a =. The nonlinear function in f(t) becomes q 2k x 2k q 2k x 2k q 3 x 3 q 2 x 2 = q 2k a 2k sin 2k (!t) q 2k a 2k sin 2k (!t) q 3 a 3 sin 3 (!t) q 2 a 2 sin 2 (!t)

5 E. Oxenhielm / Nonlinear Analysis ( ) 5 = q2k k a 2k q 2k 2 k a 2k 2 q 2 a 2 [ q 2k k a 2k q 2k k a 2k q 3 a 3 ] sin(!t) such that each l always is larger than or equal to =2 l, for integers l such that 6 l 6 k. The Lienard equation () becomes [ ] [ ][ ] [ ] ẋ q a q2k k a 2k q 2k 2 k a 2k 2 q 2 a 2 = ẏ b ([ ] [ ]) [ ] q q2k k a 2k q 3 a 2 a sin(!t): b Letting the constant part of f(t) equal to zero according to Eq. (2) gives that b = q 2k k a 2k q 2k 2 k a 2k 2 q 2 a 2 (5) so that the coecients in front of the even powers of the polynomial F(x) only have an impact on the center value b (see also Example ). Eq. (3) gives that the solutions for a and! satisfy! = ± (6) so that the dominant harmonic order is the rst one, and q 2k k a 2k q 3 a 2 q = (7) which has at most k distinct zeros in terms of a 2. For each such zero a2, we have that ± a are solutions. The approximation of the state variables as in Eq. (4) ts the solution in terms of x. To solve for y, we notice that ẏ = x from the Lienard equation (), which gives that y = b a cos(!t) (8)! since x is as in Equation (4) and! is nonzero. Notice that b still satises Eq. (5), since we would get that same equation if we assumed that the state variables were cosine functions instead of sine functions in Eq. (4). Thus, the following are the possible cases for the state variables. A. If a and! are of the same sign, x = a sin t; y = b a cos t: B. If a and! are of dierent signs x = a sin t; y = b a cos t:

6 6 E. Oxenhielm / Nonlinear Analysis ( ) Fig. 2. Numerical integration of the Lienard equation () when F(x)=x 3 :5x 2 x and the initial condition y() is positive. For each distinct zero a 2 of Eq. (7), these two cases correspond to the same limit cycle. Simulations show that the sign of the initial condition y() determines whether the trajectory follows the solution of case A or case B (see Fig. ). We thus have that for each unique x that may be approximated as in Eq. (4), there exists a unique y as in Eq. (8). Therefore, we have that there exist at most k distinct limit cycles in the Lienard equation (). In Eq. (4), we assumed that the state variables were dominated by a harmonic term of a specic order. If this assumption is not true, it is possible to increase the accuracy of the approximation of the state variables by adding higher harmonic terms to Eq. (4). By doing this, and by going through the calculations one more time, we would if the state variables were not dominated by a harmonic term of a specic order end up with other amplitudes than in the rst calculation. It is of great importance for the result to understand that this change in amplitudes does not mean that there exist additional limit cycles in the Lienard equation () it only means that the new approximation of the state variables has a higher accuracy (see also [5]). Thereby, we have proved that the maximum number k of limit cycles that exist in the Lienard equation (), depends on the degree 2k of the polynomial F(x) only. Hence, we have proved that the upper bound for H(2k ) is k. Note that the method of describing functions may be used in a similar manner as in the proof above, to nd the upper bounds for the Hilbert number in any planar polynomial vector eld. Thus, it is possible to completely solve the second part of Hilbert s 6th problem by using this approach.

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